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Chemistry

2 answers:

just olya [345]4 years ago

6 0

Answer is C)magnetic and electric fields.

yanalaym [24]4 years ago

5 0

<h2>Plasma is most likely inflence by - Option C</h2>

This plasma is most likely influenced by magnetic and electric fields. The massive positively charged ions which are influenced the lighter electrons than replace the ions to keep charge impartiality.

Therefore electric field influence the plasma. The movement of charged particles identical to a magnetic field line is not influenced. Plasma contains charged particles which are positive nuclei and negative electrons which can be molded and restricted by magnetic forces.

Alike iron filings in the closeness of a magnet, bits in the plasma will ensue magnetic field lines.

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devlian [24]

Answer:

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4 0

3 years ago

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Katyanochek1 [597]

Answer:

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4 years ago

How many grams of NaF form when 0.589 mol of HF react with excess Na2SiO3?

BartSMP [9]

Answer:

6.18 g

Explanation:

We know we will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

$M_r$ : 41.99

Na₂SiO₃ + 8HF ⟶ 2NaF + H₂SiF₆ + 3H₂O

<em>n</em>/mol: 0.58

1. Use the molar ratio of NaF:HF to calculate the moles of NaF.

$\text{Moles of NaF} = \text{0.589 mol HF} \times \frac{\text{2 mol NaF}}{\text{8 mol HF}} = \text{0.1472 mol NaF}$

2. Use the molar mass of NaF to calculate the mass of NaF.

$\text{Mass of NaF} = \text{0.1472 mol NaF} \times \frac{\text{41.99 g NaF}}{\text{1 mol NaF}} = \text{6.18 g NaF}$

8 0

3 years ago

The vapor pressure of pure acetone is 266 torr. When a non-volatile solute is added, the vapor pressure of acetone above the sol

mixas84 [53]

Answer : The mole fraction of the non-volatile solute in the solution is 0.195

Explanation :

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

Formula used :

$\frac{\Delta p}{p^o}=X_B$